Problem 305 - Calculus Volume 1

Question

For the following exercises, draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior. $$ y=x^{2} \sin (x), x=[-2 \pi, 2 \pi] $$

Step-by-Step Solution

Answer

The graph of \( y = x^2 \sin(x) \) is symmetric, with critical points at \( x = 0 \) and zero crossings at integer multiples of \( \pi \), within \([-2\pi, 2\pi]\).

Step 1: Identify the Function

The function given is \( y = x^2 \sin(x) \). This is a product of a polynomial \( x^2 \) and a trigonometric function \( \sin(x) \). You should expect the graph to reflect the oscillation of \( \sin(x) \) modulated by the parabola-like behavior of \( x^2 \).

Step 2: Determine Symmetry

The function \( y = x^2 \sin(x) \) is **even** because \( y(-x) = (-x)^2 \sin(-x) = x^2 (-\sin(x)) = x^2 \sin(x) \). Therefore, the graph will be symmetric with respect to the y-axis.

Step 3: Find Critical Points

Set the derivative of the function equal to zero to find critical points. Differentiate using the product rule:\( y' = \frac{d}{dx}(x^2 \sin(x)) = 2x \sin(x) + x^2 \cos(x) \).Set \( y' = 0 \): \( 2x \sin(x) + x^2 \cos(x) = 0 \). Factoring out \( x \), we get: \( x(2 \sin(x) + x \cos(x)) = 0 \). So, critical points occur when \( x = 0 \) or \( 2 \sin(x) + x \cos(x) = 0 \).

Step 4: Determine Inflection Points

Find the second derivative to determine inflection points. Differentiate the first derivative:\( y'' = \frac{d}{dx}(2x \sin(x) + x^2 \cos(x)) = 2\sin(x) + 2x\cos(x) + 2x\cos(x) - x^2\sin(x) \).Simplifying, we have: \( y'' = 2\sin(x) + 4x\cos(x) - x^2\sin(x) \).Set \( y'' = 0 \) to find points of inflection. Solve the equation to find the x-values where this occurs.

Step 5: Identify Asymptotic Behavior

Examine the behavior as \( x \) approaches positive and negative infinity or any vertical asymptotes. Since the domain given is \([-2\pi, 2\pi]\), there's no vertical asymptote; however, at the boundaries, the graph follows the behavior of \( y = x^2 \sin(x) \), which will not tend to infinity as sin(x) oscillates between -1 and 1. The graph flattens as \( x \to 0 \) because \( \sin(x) \to 0 \).

Step 6: Sketch the Graph

With the critical points, inflection points, and behavior identified, you can sketch a rough graph. Start by plotting points, especially at the zero crossings of \( \sin(x) \) (i.e., \( x = n\pi \), where \( n \) is an integer). Ensure symmetry around \( y = 0 \) and respect the boundaries \( x = -2\pi \) and \( x = 2\pi \). Note the modulation effect of \( \sin(x) \) on \( x^2 \).

Key Concepts

Critical Points

Critical points of a function are where the derivative is zero or undefined, leading to possible locations of local maxima, minima, or flat points upon the curve. For the function given, \( y = x^2 \sin(x) \), we found the critical points by differentiating and setting the derivative equal to zero. Using the product rule, we have:

• \( y' = 2x \sin(x) + x^2 \cos(x) \)

Setting \( y' = 0 \), we factor out \( x \), which gives:

• \( x(2 \sin(x) + x \cos(x)) = 0 \)

From this, critical points occur when \( x = 0 \) or \( 2 \sin(x) + x \cos(x) = 0 \). These points indicate positions on the graph where it might change direction, flatten, or peak. Ensure each solution is computed either numerically or graphically for their specific coordinates.

Inflection Points

Inflection points indicate where the concavity of the function changes, represented where the second derivative equals zero. This transformation from concave up to concave down or vice versa helps illustrate the nature of the curve's bending. For the function \( y = x^2 \sin(x) \), the second derivative is computed as:

• \( y'' = 2\sin(x) + 4x\cos(x) - x^2\sin(x) \)

Setting \( y'' = 0 \) allows you to find the exact points where this occurs. Solving for these x-values will grant insight into sections of the graph transitioning from peaks to troughs or vice versa.

It's crucial to interpret these inflection points visually as part of understanding how curvature and wave-like modulations affect the overall appearance of the graph.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as it approaches certain bounds or infinity, offering a glimpse into its long-term tendencies. For the function \( y = x^2 \sin(x) \), we need to look at the given domain \([-2\pi, 2\pi]\) as it determines any specific asymptotic characteristics.

• The oscillation of \( \sin(x) \) ensures the function stays bounded within -1 and 1.
• As \( x \) approaches 0, \( \sin(x) \) drives the product towards zero.
• Due to no horizontal or vertical asymptotes, focus mainly on the edges of the domain in relation to \( x^2 \), which grows quadratically but is tempered by the sine term.

This behavior underscores how trigonometric functions confine polynomial expansion over the interval, resulting in a bounded yet oscillating graph.

Trigonometric Functions

Trigonometric functions are fundamental in determining periodic behavior and oscillating patterns. In this exercise, the function \( y = x^2 \sin(x) \) incorporates \( \sin(x) \) which offers clear rhythmic oscillations.

• \( \sin(x) \) is periodic with a period of \( 2\pi \).
• The wave-like nature results in a repeated pattern that crosses the x-axis at multiples of \( \pi \).
• Oscillations from \( \sin(x) \) modulate the parabola \( x^2 \), which determines the amplitude of the wave based on the distance away from the origin.

Understanding trigonometric functions like \( \sin(x) \) is crucial for grasping the overall modulation and behavior of the graph, informing points of intersection and midpoint adjustments along the plotted range.